Factorise $x^n + 1$

Question

Is there a way to factorise $x^n + 1$

I thought of doing it like this: $$x^n +1 = (x+1)(x^{n-1} - x^{n-2} + \cdots - x + 1)$$ but can't seem to get anywhere using this method.

Answer 1

Write $$ x^n+1 = \frac{x^{2n}-1}{x^n-1} $$ The factorization over $\mathbb Q$ of the polynomials in the fraction are given by cyclotomic polynomials.

Answer 2

Factorization amounts to root finding.

$$x^n=-1\iff x=e^{i(2k+1)\pi/n}$$ and the factors are

$$x-e^{i(2k+1)\pi/n}$$ for $k=0,\cdots n-1$ (modulo $n$).

If you want a factorization with real coefficients only, then consider the roots in conjugate pairs.

$$(x-e^{i(2k+1)\pi/n})(x-e^{-i(2k+1)\pi/n})=x^2-2\cos\left(\frac{(2k+1)\pi}{n}\right)x+1.$$

Answer 3

If $n$ is even then the graph of $y=x^n+1$ is always above the $x$ axis so $x^n+1=0$ has no real roots.

If $n$ is odd then the graph of $y=x^n+1$ crosses the $x$ axis at $x=-1$ and cannot cross the $x$ axis anywhere else because $x^n+1$ is monotonically increasing. So $x^n+1=0$ has a single real root at $x=-1$, and so $x+1$ is a factor of $x^n+1$. In fact, for odd $n$:

$x^n+1=(x+1)(x^{n-1}-x^{n-2}+x^{n-3}-x^{n-4}+\dots-x+1)$